NCERT Solutions for Class 10 Maths Unit 8
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NCERT Solutions for Class 10 Maths Unit 8
Introduction to Trigonometry Class 10
Unit 8 Introduction to Trigonometry Exercise 8.1, 8.2, 8.3, 8.4 Solutions
Exercise 8.1 : Solutions of Questions on Page Number : 181 Q1 :
In ABC right angled at B, AB = 24 cm, BC = 7 m. Determine (i) sin A, cos A (ii) sin C, cos C
Answer : Applying Pythagoras theorem for ABC, we obtain AC2 = AB2 + BC2 = (24 cm)2 + (7 cm)2 = (576 + 49) cm2 = 625 cm2
AC =
cm = 25 cm
(i) sin A =
cos A = (ii)
sin C =
cos C =
Q2 : In the given figure find tan P - cot R
Answer : Applying Pythagoras theorem for PQR, we obtain PR2 = PQ2 + QR2 (13 cm)2 = (12 cm)2 + QR2 169 cm2 = 144 cm2 + QR2 25 cm2 = QR2 QR = 5 cm
tan P - cot R =
Q3 :
If sin A = , calculate cos A and tan A. Answer : Let ABC be a right-angled triangle, right-angled at point B.
Given that,
Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer. Applying Pythagoras theorem in ABC, we obtain AC2 = AB2 + BC2 (4k)2 = AB2 + (3k)2 16k 2 - 9k 2 = AB2 7k 2 = AB2 AB =
Q4 : Given 15 cot A = 8. Find sin A and sec A Answer : Consider a right-angled triangle, right-angled at B.
It is given that,
cot A = Let AB be 8k.Therefore, BC will be 15k, where k is a positive integer. Applying Pythagoras theorem in ABC, we obtain AC2 = AB2 + BC2 = (8k)2 + (15k)2 = 64k2 + 225k2 = 289k2 AC = 17k
Q5 : Given sec = , calculate all other trigonometric ratios. Answer : Consider a right-angle triangle ABC, right-angled at point B.
If AC is 13k, AB will be 12k, where k is a positive integer. Applying Pythagoras theorem in ABC, we obtain (AC)2 = (AB)2 + (BC)2 (13k)2 = (12k)2 + (BC)2 169k2 = 144k2 + BC2 25k2 = BC2 BC = 5k
Q6 :
If A and B are acute angles such that cos A = cos B, then show that A = B.
Answer :
Let us consider a triangle ABC in which CD AB.
It is given that cos A = cos B
... (1)
We have to prove A = B. To prove this, let us extend AC to P such that BC = CP.
From equation (1), we obtain
By using the converse of B.P.T, CD||BP
ACD = CPB (Corresponding angles) ... (3) And, BCD = CBP (Alternate interior angles) ... (4)
By construction, we have BC = CP.
obtainCBP = CPB (Angle opposite to equal sides of a triangle) ... (5) From equations (3), (4), and (5), we ACD = BCD ... (6)
ACD = BCD [Using equation (6)]
CDA = CDB [Both 90?]
Therefore, the remaining angles should be equal.
CAD = CBD A = B
Alternatively,
Let us consider a triangle ABC in which CD AB.
It is given that, cos A = cos B
Let
AD = k BD ... (1)
And, AC = k BC ... (2)
Using Pythagoras theorem for triangles CAD and CBD, we obtain
CD2 = AC2 - AD2 ... (3)
And, CD2 = BC2 - BD2 ... (4)
From equations (3) and (4), we obtain
AC2 - AD2 = BC2
(k BC)2 - (k BD)2 = BC2 - BD2 k2 (BC2 - BD2) = BC2 - BD2 k2 = 1 k=1
- BD2
A = B(Angles opposite to equal sides of a triangle)
Q7 : If cot = , evaluate
(i)
(ii) cot2
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